Show that $x-1$ is a factor of $P(x)$ where $P(x^5)+xQ(x^5)+x^2R(x^5)=big(x^4+x^3+x^2+x+1big)S(x)$ [duplicate]

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  • If $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$

    2 answers



For the problem below, my first step was to rewrite the right side as



$$big(x^4+x^3+x^2+x+1big)S(x)=fracx^5-1x-1S(x)$$



From there I can isolate $P(x^5)$ which gives me



$$P(x^5)=dfracS(x)(x^5-1)-x(x-1)big(Q(x^5)-xR(x^5)big)x-1$$



I'm not sure how to move forward. To get $P(x)$ should I rewrite all the $x^5$ as $x$ and take the $5^th$ root of each $x$? The questions prior used the idea of roots of unity so perhaps I can use that somehow.



Thanks



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polynomials proof-writing contest-math




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marked as duplicate by greedoid, Key Flex, Batominovski, Dave, Anurag A yesterday


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    up vote
    -1
    down vote

    favorite
    1













    This question already has an answer here:



    • If $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$

      2 answers



    For the problem below, my first step was to rewrite the right side as



    $$big(x^4+x^3+x^2+x+1big)S(x)=fracx^5-1x-1S(x)$$



    From there I can isolate $P(x^5)$ which gives me



    $$P(x^5)=dfracS(x)(x^5-1)-x(x-1)big(Q(x^5)-xR(x^5)big)x-1$$



    I'm not sure how to move forward. To get $P(x)$ should I rewrite all the $x^5$ as $x$ and take the $5^th$ root of each $x$? The questions prior used the idea of roots of unity so perhaps I can use that somehow.



    Thanks



    enter image description here




    polynomials proof-writing contest-math




    share|cite|improve this question













    marked as duplicate by greedoid, Key Flex, Batominovski, Dave, Anurag A yesterday


    This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
















      up vote
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      up vote
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      This question already has an answer here:



      • If $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$

        2 answers



      For the problem below, my first step was to rewrite the right side as



      $$big(x^4+x^3+x^2+x+1big)S(x)=fracx^5-1x-1S(x)$$



      From there I can isolate $P(x^5)$ which gives me



      $$P(x^5)=dfracS(x)(x^5-1)-x(x-1)big(Q(x^5)-xR(x^5)big)x-1$$



      I'm not sure how to move forward. To get $P(x)$ should I rewrite all the $x^5$ as $x$ and take the $5^th$ root of each $x$? The questions prior used the idea of roots of unity so perhaps I can use that somehow.



      Thanks



      enter image description here




      polynomials proof-writing contest-math




      share|cite|improve this question














      This question already has an answer here:



      • If $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$

        2 answers



      For the problem below, my first step was to rewrite the right side as



      $$big(x^4+x^3+x^2+x+1big)S(x)=fracx^5-1x-1S(x)$$



      From there I can isolate $P(x^5)$ which gives me



      $$P(x^5)=dfracS(x)(x^5-1)-x(x-1)big(Q(x^5)-xR(x^5)big)x-1$$



      I'm not sure how to move forward. To get $P(x)$ should I rewrite all the $x^5$ as $x$ and take the $5^th$ root of each $x$? The questions prior used the idea of roots of unity so perhaps I can use that somehow.



      Thanks



      enter image description here





      This question already has an answer here:



      • If $P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x)$ , then prove that $P(x)$ is divisible by $x-1$

        2 answers





      polynomials proof-writing contest-math





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      edited 20 hours ago









      greedoid

      26.1k93373




      26.1k93373









      asked yesterday









      john fowles

      1,052716




      1,052716




      marked as duplicate by greedoid, Key Flex, Batominovski, Dave, Anurag A yesterday


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






      marked as duplicate by greedoid, Key Flex, Batominovski, Dave, Anurag A yesterday


      This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






















          1 Answer
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          Hint: You know that if $P(1)=0$, then $x-1$ is a factor of $P$. Try plugging in special values of $x$ into the given functional equation. Try certain algebraic numbers with nice properties, such as, for example, the roots of unity.






          share|cite|improve this answer





















          • In the other answer to this same question, they say that "Plugging in $omega, omega^2, cdots,omega^4$ in the equations successively we get that $P(1)+xQ(1)+x^2 R(1) = 0$ for four distinct complex numbers. Hence it must be identically 0." They then conclude that $P(1),Q(1),R(1)=0$. Why is it that $4$ different values of $x$ all giving $0$ implies that the $3$ polynomials equal $0$ at $1$? Thanks for the response
            – john fowles
            yesterday











          • Because quadratic function can not cut a horizontal line more than twice.
            – greedoid
            yesterday










          • @greedoid So we could've made the same conclusion by just showing that $3$ different values of $x$ make the right side equal to $0$
            – john fowles
            yesterday

















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote













          Hint: You know that if $P(1)=0$, then $x-1$ is a factor of $P$. Try plugging in special values of $x$ into the given functional equation. Try certain algebraic numbers with nice properties, such as, for example, the roots of unity.






          share|cite|improve this answer





















          • In the other answer to this same question, they say that "Plugging in $omega, omega^2, cdots,omega^4$ in the equations successively we get that $P(1)+xQ(1)+x^2 R(1) = 0$ for four distinct complex numbers. Hence it must be identically 0." They then conclude that $P(1),Q(1),R(1)=0$. Why is it that $4$ different values of $x$ all giving $0$ implies that the $3$ polynomials equal $0$ at $1$? Thanks for the response
            – john fowles
            yesterday











          • Because quadratic function can not cut a horizontal line more than twice.
            – greedoid
            yesterday










          • @greedoid So we could've made the same conclusion by just showing that $3$ different values of $x$ make the right side equal to $0$
            – john fowles
            yesterday














          up vote
          1
          down vote













          Hint: You know that if $P(1)=0$, then $x-1$ is a factor of $P$. Try plugging in special values of $x$ into the given functional equation. Try certain algebraic numbers with nice properties, such as, for example, the roots of unity.






          share|cite|improve this answer





















          • In the other answer to this same question, they say that "Plugging in $omega, omega^2, cdots,omega^4$ in the equations successively we get that $P(1)+xQ(1)+x^2 R(1) = 0$ for four distinct complex numbers. Hence it must be identically 0." They then conclude that $P(1),Q(1),R(1)=0$. Why is it that $4$ different values of $x$ all giving $0$ implies that the $3$ polynomials equal $0$ at $1$? Thanks for the response
            – john fowles
            yesterday











          • Because quadratic function can not cut a horizontal line more than twice.
            – greedoid
            yesterday










          • @greedoid So we could've made the same conclusion by just showing that $3$ different values of $x$ make the right side equal to $0$
            – john fowles
            yesterday












          up vote
          1
          down vote










          up vote
          1
          down vote









          Hint: You know that if $P(1)=0$, then $x-1$ is a factor of $P$. Try plugging in special values of $x$ into the given functional equation. Try certain algebraic numbers with nice properties, such as, for example, the roots of unity.






          share|cite|improve this answer













          Hint: You know that if $P(1)=0$, then $x-1$ is a factor of $P$. Try plugging in special values of $x$ into the given functional equation. Try certain algebraic numbers with nice properties, such as, for example, the roots of unity.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered yesterday









          Frpzzd

          16.9k63489




          16.9k63489











          • In the other answer to this same question, they say that "Plugging in $omega, omega^2, cdots,omega^4$ in the equations successively we get that $P(1)+xQ(1)+x^2 R(1) = 0$ for four distinct complex numbers. Hence it must be identically 0." They then conclude that $P(1),Q(1),R(1)=0$. Why is it that $4$ different values of $x$ all giving $0$ implies that the $3$ polynomials equal $0$ at $1$? Thanks for the response
            – john fowles
            yesterday











          • Because quadratic function can not cut a horizontal line more than twice.
            – greedoid
            yesterday










          • @greedoid So we could've made the same conclusion by just showing that $3$ different values of $x$ make the right side equal to $0$
            – john fowles
            yesterday
















          • In the other answer to this same question, they say that "Plugging in $omega, omega^2, cdots,omega^4$ in the equations successively we get that $P(1)+xQ(1)+x^2 R(1) = 0$ for four distinct complex numbers. Hence it must be identically 0." They then conclude that $P(1),Q(1),R(1)=0$. Why is it that $4$ different values of $x$ all giving $0$ implies that the $3$ polynomials equal $0$ at $1$? Thanks for the response
            – john fowles
            yesterday











          • Because quadratic function can not cut a horizontal line more than twice.
            – greedoid
            yesterday










          • @greedoid So we could've made the same conclusion by just showing that $3$ different values of $x$ make the right side equal to $0$
            – john fowles
            yesterday















          In the other answer to this same question, they say that "Plugging in $omega, omega^2, cdots,omega^4$ in the equations successively we get that $P(1)+xQ(1)+x^2 R(1) = 0$ for four distinct complex numbers. Hence it must be identically 0." They then conclude that $P(1),Q(1),R(1)=0$. Why is it that $4$ different values of $x$ all giving $0$ implies that the $3$ polynomials equal $0$ at $1$? Thanks for the response
          – john fowles
          yesterday





          In the other answer to this same question, they say that "Plugging in $omega, omega^2, cdots,omega^4$ in the equations successively we get that $P(1)+xQ(1)+x^2 R(1) = 0$ for four distinct complex numbers. Hence it must be identically 0." They then conclude that $P(1),Q(1),R(1)=0$. Why is it that $4$ different values of $x$ all giving $0$ implies that the $3$ polynomials equal $0$ at $1$? Thanks for the response
          – john fowles
          yesterday













          Because quadratic function can not cut a horizontal line more than twice.
          – greedoid
          yesterday




          Because quadratic function can not cut a horizontal line more than twice.
          – greedoid
          yesterday












          @greedoid So we could've made the same conclusion by just showing that $3$ different values of $x$ make the right side equal to $0$
          – john fowles
          yesterday




          @greedoid So we could've made the same conclusion by just showing that $3$ different values of $x$ make the right side equal to $0$
          – john fowles
          yesterday


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